Question 1

In the trick given above, what is the quotient when you divide by 9? Is there a relationship between the two numbers and the quotient?

Accepted Answer

When the difference between the two-digit number ab and its reverse ba is calculated, the difference is 9(b - a). Dividing this difference by 9 gives the quotient (b - a). Thus, the quotient is the difference between the digits of the original number. — Let the two-digit number be ab = 10a + b. Its reverse is ba = 10b + a. The difference is (10b + a) - (10a + b) = 9(b - a). Dividing by 9 gives (b - a). Hence, the quotient is the difference between the digits.

Question 2

In the trick given above, instead of finding the difference of the two 2-digit numbers, find their sum. What will happen? For example:
- We start with 31. After reversing we get 13. Adding 31 and 13, we get 44.
- We start with 28. After reversing we get 82. Adding 28 and 82, we get 110.
- We start with 12. After reversing we get 21. Adding 12 and 21, we get 33.

Observe that all these numbers are divisible by 11. Is this always true? Can we justify this claim using algebra?

Accepted Answer

Let the two-digit number be ab = 10a + b. Its reverse is ba = 10b + a. Their sum is (10a + b) + (10b + a) = 11(a + b). Since 11(a + b) is always divisible by 11, the sum of a two-digit number and its reverse is always divisible by 11. — Sum = (10a + b) + (10b + a) = 11(a + b). Since 11 is a factor, the sum is divisible by 11 for any digits a and b.

Question 3

Consider any 3-digit number, say abc (100a + 10b + c). Make two other 3-digit numbers from these digits by cycling these digits around, yielding bca and cab. Now add the three numbers. Using algebra, justify that the sum is always divisible by 37. Will it also always be divisible by 3? [Hint: Look at some multiples of 37.]

Accepted Answer

Let the 3-digit number be abc = 100a + 10b + c.
The cyclic numbers are:
- bca = 100b + 10c + a
- cab = 100c + 10a + b
Sum = abc + bca + cab = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b) = 111(a + b + c).
Since 111 = 3 × 37, the sum is divisible by 37 and also by 3. — Sum = 111(a + b + c). Because 111 is divisible by 37 and 3, the sum is divisible by both 37 and 3 for any digits a, b, c.

Question 4

Consider any 3-digit number, say abc. Make it a 6-digit number by repeating the digits, that is abcabc. Divide this number by 7, then by 11, and finally by 13. What do you get? Try this with other numbers. Figure out why it works. [Hint: Multiply 7, 11 and 13.]

Accepted Answer

Let the 3-digit number be abc = 100a + 10b + c.
The 6-digit number formed by repeating abc is abcabc = 1000 × abc + abc = 1001 × abc.
Since 1001 = 7 × 11 × 13, dividing abcabc by 7, 11, and 13 successively will give abc.
This works for any 3-digit number abc. — abcabc = abc × 1001 = abc × (7 × 11 × 13). Dividing by 7, then 11, then 13 returns abc.

Question 5

There are 3 shrines, each with a magical pond in the front. If anyone dips flowers into these magical ponds, the number of flowers doubles. A person has some flowers. He dips them all in the first pond and then places some flowers in shrine 1. Next, he dips the remaining flowers in the second pond and places some flowers in shrine 2. Finally, he dips the remaining flowers in the third pond and then places them all in shrine 3. If he placed an equal number of flowers in each shrine, how many flowers did he start with? How many flowers did he place in each shrine?

Accepted Answer

Let the number of flowers placed in each shrine be x.
Let the initial number of flowers be N.
Step 1: Dip all N flowers in first pond → number doubles to 2N.
Place x flowers in shrine 1 → remaining flowers = 2N - x.
Step 2: Dip remaining flowers in second pond → number doubles to 2(2N - x) = 4N - 2x.
Place x flowers in shrine 2 → remaining flowers = 4N - 2x - x = 4N - 3x.
Step 3: Dip remaining flowers in third pond → number doubles to 2(4N - 3x) = 8N - 6x.
Place x flowers in shrine 3 → remaining flowers = 8N - 6x - x = 8N - 7x.
Since all flowers are placed, remaining flowers after shrine 3 is zero:
8N - 7x = 0 → 8N = 7x → x = (8/7)N.
But x is the number placed in each shrine, so it must be less than or equal to total flowers after doubling.
From step 1, x ≤ 2N.
From x = (8/7)N, x > N, so N must be divisible by 7 to get integer values.
Choose N = 7k, then x = 8k.
Since x > 2N = 14k, contradiction unless k=0.
Re-examining, since x = (8/7)N, to have integer values, let N = 7.
Then x = 8.
Check:
Start with 7 flowers.
After first pond: 14 flowers.
Place 8 in shrine 1 → remaining 6.
Second pond: 12 flowers.
Place 8 in shrine 2 → remaining 4.
Third pond: 8 flowers.
Place 8 in shrine 3 → remaining 0.
Thus, he started with 7 flowers and placed 8 flowers in each shrine. — Using algebraic expressions for each step and the condition that equal flowers are placed in each shrine, we solve for initial flowers and flowers placed per shrine. The solution is 7 flowers initially and 8 flowers placed in each shrine.

Question 6

A farm has some horses and hens. The total number of heads of these animals is 55 and the total number of legs is 150. How many horses and how many hens are on the farm?

Can you solve this without letter-numbers?

[Hint: If all the 55 animals were hens, then how many legs would there be? Using the difference between this number and 150, can you find the number of horses?]

Accepted Answer

Let the number of hens be h and horses be H.
Total heads: h + H = 55.
Total legs: 2h + 4H = 150.
Without letters:
If all 55 were hens, total legs = 55 × 2 = 110.
Actual legs = 150.
Difference = 150 - 110 = 40.
Each horse has 2 more legs than a hen.
Number of horses = 40 / 2 = 20.
Number of hens = 55 - 20 = 35. — Using the hint, calculate legs if all were hens, find difference, divide by 2 to get horses, subtract to get hens.

Question 7

A mother is 5 times her daughter's age. In 6 years' time, the mother will be 3 times her daughter's age. How old is the daughter now?

Accepted Answer

Let daughter's current age be x.
Mother's current age = 5x.
After 6 years:
Daughter's age = x + 6.
Mother's age = 5x + 6.
Given: 5x + 6 = 3(x + 6).
5x + 6 = 3x + 18.
5x - 3x = 18 - 6.
2x = 12.
x = 6.
Daughter is 6 years old now. — Set up equations based on given ratios and solve for daughter's age.

Question 8

Two friends, Gauri and Naina, are cowherds. One day, they pass each other on the road with their cows. Gauri says to Naina, "You have twice as many cows as I do". Naina says, "That's true, but if I gave you three of my cows, we would each have the same number of cows". How many cows do Gauri and Naina have?

Accepted Answer

Let Gauri have x cows.
Naina has 2x cows.
If Naina gives 3 cows to Gauri:
Gauri's cows = x + 3.
Naina's cows = 2x - 3.
Given they have the same number:
x + 3 = 2x - 3.
3 + 3 = 2x - x.
6 = x.
Gauri has 6 cows.
Naina has 12 cows. — Translate the problem into equations and solve for x.

Question 9

I run a small dosa cart and my expenses are as follows:
- Rent for the dosa cart is ₹5000 per day.
- The cost of making one dosa (including all the ingredients and fuel) is ₹10.

(i) If I can sell 100 dosas a day, what should be the selling price of my dosa to make a profit of ₹2000?

(ii) If my customers are willing to pay only ₹50 for a dosa, how many dosas should I aim to sell in a day to make a profit of ₹2000?

Accepted Answer

(i) Let selling price per dosa be p.
Number of dosas sold = 100.
Total cost = Rent + Cost of dosas = 5000 + 10 × 100 = 5000 + 1000 = ₹6000.
Total revenue = 100 × p.
Profit = Revenue - Cost = 2000.
So, 100p - 6000 = 2000 → 100p = 8000 → p = ₹80.

(ii) Selling price p = ₹50.
Let number of dosas sold be n.
Total cost = 5000 + 10n.
Total revenue = 50n.
Profit = Revenue - Cost = 2000.
So, 50n - (5000 + 10n) = 2000 → 50n - 5000 - 10n = 2000 → 40n = 7000 → n = 175 dosas. — Use profit formula: Profit = Revenue - Cost. Set up equations for each part and solve for price or quantity.

Question 10

Evaluate the following sequence of fractions:

$$
\frac{1}{3}, \frac{(1 + 3)}{(5 + 7)}, \frac{(1 + 3 + 5)}{(7 + 9 + 11)}
$$

What do you observe? Can you explain why this happens?

[Hint: Recall what you know about the sum of the first n odd numbers.]

Accepted Answer

Calculate each fraction:
1/3 = 0.333...
(1+3)/(5+7) = 4/12 = 1/3 = 0.333...
(1+3+5)/(7+9+11) = 9/27 = 1/3 = 0.333...
Observation: All fractions equal 1/3.
Explanation:
Sum of first n odd numbers = n².
Numerator sums are sums of first n odd numbers:
- 1 = 1²
- 1+3 = 2² = 4
- 1+3+5 = 3² = 9
Denominator sums are sums of next n odd numbers starting from (2n+1):
- For n=1: 3rd odd number is 5, but here denominator is 3,5+7,7+9+11, so the denominator sums to 3 × n².
Actually, denominator sums are 3 times numerator sums.
Hence, fraction = numerator / denominator = n² / (3 n²) = 1/3. — Using the formula for sum of first n odd numbers and recognizing denominator sums as triple numerator sums, the fraction simplifies to 1/3.

Question 11

Karim and the Genie

Karim was taking a nap under a tree. He had a dream about a magical lamp and a genie. He heard a voice saying, "I have come to serve you, Oh master". He woke up and to his surprise, it was a genie!

"Do you want to make money?", asked the genie. Karim nodded dumbly in bewilderment. The genie continued, "Do you see the banyan tree over there? All you have to do is go around it once. The money in your pocket will double".

Karim immediately started towards the tree, only to be stopped by the genie. "One moment!", said the genie. "Since I am bringing you great riches, you should share some of your gains with me. You must give me 8 coins each time you go around the tree."

Thinking that was a trifling amount, Karim readily agreed.

He went around the tree once. Just as the genie had said, the number of coins in his pocket doubled! He gave 8 coins to the genie. He made another round. Again the number of coins doubled. He gave 8 more coins to the genie. He went around the tree for the third time. The number of coins doubled again, but to his horror, he was left with only 8 coins, exactly the number of coins he owed the genie!

As Karim began to wonder how the genie tricked him, the genie let out a loud laugh and disappeared.

(i) How many coins did Karim initially have?
(ii) For what cost per round should Karim agree to the deal, if he wants to increase the number of coins he has?
(iii) Through its magical powers, the genie knows the number of coins that Karim has. How should the genie set the cost per round so that it gets all of Karim's coins?

Accepted Answer

(i) Let initial coins be x.
After first round: coins double to 2x, then give 8 to genie → coins left = 2x - 8.
After second round: coins double to 2(2x - 8) = 4x - 16, give 8 → coins left = 4x - 24.
After third round: coins double to 2(4x - 24) = 8x - 48, give 8 → coins left = 8x - 56.
Given coins left after third round = 8.
So, 8x - 56 = 8 → 8x = 64 → x = 8.
Karim initially had 8 coins.

(ii) Let cost per round be c.
After n rounds, coins = (2^n) * x - c * (2^{n} - 1).
For coins to increase after each round, net coins after round n > coins after round (n-1).
This implies c < x.
So, cost per round should be less than initial coins to increase coins.

(iii) Genie wants to get all coins.
Set cost c such that after some rounds coins left = 0.
From formula: coins after n rounds = (2^n) * x - c * (2^{n} - 1) = 0.
Solving for c: c = (2^n * x) / (2^{n} - 1).
Genie can set c accordingly to get all coins after n rounds. — Using algebraic expressions for coins after each round and given conditions, solve for initial coins and cost per round.

Question 12

What is algebra and why is it important in solving problems where exact numbers are unknown?

Accepted Answer

Algebra is a branch of mathematics that uses symbols, usually letters, to represent numbers in equations and formulas. It is important because it allows us to represent unknown numbers with variables and solve problems even when exact numbers are not known. For example, algebra helps in formulating general rules and solving unknowns in science and economics. — Algebra uses letters to represent numbers which may be unknown or variable. This symbolic representation helps in solving equations and formulating general rules applicable in various fields. For instance, in economics, algebra helps calculate profit or loss without knowing exact amounts initially.

Question 13

Which of the following is an algebraic expression?

Accepted Answer

7x - 4 — An algebraic expression contains variables and constants combined using arithmetic operations. Among the options, only '7x - 4' contains a variable 'x' along with constants and operations, making it an algebraic expression.

Question 14

Identify the coefficient and degree of the term $7x^2$.

Accepted Answer

The coefficient of the term $7x^2$ is 7, which is the numerical factor multiplying the variable. The degree of the term is 2, which is the exponent of the variable $x$. — In the term $7x^2$, 7 is the coefficient because it multiplies the variable part. The exponent 2 indicates the degree of the term, representing the power to which the variable is raised.

Question 15

What is the difference between a monomial, binomial, and polynomial?

Accepted Answer

A monomial is an algebraic expression with one term, for example, $5x$. A binomial has two terms, like $3x + 4$. A polynomial has more than two terms, such as $x^2 + 3x + 2$. — Monomials contain a single term, binomials have exactly two terms, and polynomials have three or more terms. Understanding these helps classify algebraic expressions based on the number of terms.

Algebra Play

Algebra Play — Study Notes

6.1 Algebra Play

6.2 Thinking about ‘Think of a Number’ Tricks

6.3 Number Pyramids

Practice Questions — Algebra Play

All 7 Chapters in Ganita Prakash Part-II